\textbf{\Huge Problems}
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% Problem 1
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\section{Consider the following statements:}
$(1+2)^2-1^2=2^3$\\
$(1+2+3)^2-(1+2)^2=3^3$\\
$(1+2+3+4)^2-(1+2+3)^2=4^3$\\
\begin{enumerate}[(a)]
\item Based on the three statements given above, what is the next statement suggested by these?
\item What conjecture is suggested by these statements?
\item Verify the conjecture in (b).
\end{enumerate}
\textbf{\Large(a)}\\
We see that the next statement would be: $(1+2+3+4+5)^2-(1+2+3+4)^2=5^3$.\\
\textbf{\Large(b)}\\
The conjecture is $(1+2+3+4+5+...+n)^2-(1+2+3+4+...+n-1)^2=n^3$.\\
\textbf{\Large(c)}\\
Base case:\\
Let n = 2: $(1+2)^2-(1)^2 = 2^3$\\
$9-1 = 8$ so it is true.\\
We know from Theorem 6.3 that p(n) can be rewritten as
\begin{equation}
\left(\frac{n(n+1)}{2}\right)^2 - \left(\frac{n-1(n)}{2}\right)^2 = n^3
\end{equation}
We assume p(k).
Let p(k + 1):\\
\begin{equation}
(k+1)^3 = \left(\frac{(k+1)(k+1+1)}{2}\right)^2 - \left(\frac{(k+1-1)(k+1+1-1)}{2}\right)^2 
\end{equation}
\begin{equation}
 = \left(\frac{k^2+3k+2}{2}\right)^2 - \left(\frac{k^2+k}{2}\right)^2 
\end{equation}
\begin{equation}
 = \frac{k^4}{4} + \frac{3k^3}{2} + \frac{13k^2}{4} + 3k + 1 - \left(\frac{k^4}{4} + \frac{k^3}{2} + \frac{k^2}{4}\right)
\end{equation}
\begin{equation}
 = \frac{2k^3}{2} + \frac{12k^2}{4} + 3k + 1
\end{equation}
\begin{equation}
 = k^3 + 3k^2 + 3k + 1
\end{equation}
\begin{equation}
 = (k+1)^3
\end{equation}
Our conjecture was correct.\includegraphics[scale=0.70]{billeder/xzibit}

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% Problem 2
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\section{By an ordered partition of an integer $n\geq 2$ is meant a sequence of positive integers whose sum is n.}
For example, the ordered partitions of 3 are 3, 1 + 2, 2 + 1, 1 + 1 + 1.
\begin{enumerate}[(a)]
\item Determine the ordered partitions of 4.
\item Determine the ordered partitions of 5.
\item Make a conjecture concerning the number of ordered partitions of an integer $n\geq 2$
\end{enumerate}
\textbf{\Large(a)}\\
The ordered partitions of 4 are as following: \\
4,
 3 + 1, 1 + 3,
  2 + 2, 2 + 1 + 1, 1 + 2 + 1, 1 + 1 + 2,
   1 + 1 + 1 + 1\\
\textbf{\Large(b)}\\
The ordered partitions of 5 are as following: \\
5,
 4 + 1, 1 + 4,
  3 + 2, 3 + 1 + 1, 2 + 3, 1 + 1 + 3, 1 + 3 + 1
   2 + 2 + 1, 2 + 1 + 2, 1 + 2 + 2, 2 + 1 + 1 + 1, 1 + 2 + 1 + 1, 1 + 1 + 2 + 1, 1 + 1 + 1 + 2,
    1 + 1 + 1 + 1 + 1\\
\textbf{\Large(c)}\\
We see that 3 has 4 ordered partitions, 4 has 8 ordered partitions and 5 has 16 ordered partitions. This can also be written as $2^2$, $2^3$ and $2^4$. This leads to the conjecture that the number of ordered partitions of an integer $n\geq 2$ is $2^{n-1}$.\includegraphics[scale=0.70]{billeder/xzibit}


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% Problem 3
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\section{Express the following quantified statement in symbols:}
For every odd integer n, the integer 3n+1 is even.\\
$\forall n\in \mathbb{Z} : 2\nmid n , 2|3n+1$\\
\textbf{Part (b)} \\
Prove that the statement is true.\\
For the statement to be true if must be of the form 2a. We substitute n with the odd form k+1 reduce the statement:\\
$3(2k+1)+1 = 6k +3 +1 = 2(3k+2)$. It ends up being divisible by 2 and therefore the statement is true.\includegraphics[scale=0.70]{billeder/xzibit}


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% Problem 4
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\section{Express the following quantified statement in symbols:}
There exists a positive integer n such that $3n+2^{n-2}$is odd.\\
$\exists n \in \mathbb{Z} : 2|n, 2\nmid 3n+2^{n-2}$\\
\textbf{Part (b)} \\
Prove that the statement is true.\\
For the statement to be true we need to find just one example because it states that there exists at least one:\\
Consider $n=2$
$3*2+2^{2-2}=6+2^0=6+1=7$ Therefore the statement is true because we found one that renders it true..\includegraphics[scale=0.70]{billeder/xzibit}


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% Problem 5
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\section{Prove or disprove: The sum of every five consecutive integers is divisable by 5 and the sum of no six consecutive integers is divisable by 6}
We start with the sum of every five consectutive integers whether it is divisable by 5:\\
$(n)+(n+1)+(n+2)+(n+3)+(n+4) = 5n+10 = 5(n+2)$\\
This is always divisible with 5 so it is true. \includegraphics[scale=0.70]{billeder/xzibit}\\
We look at the sum of six consecutive integers whether it is divisable by 6:\\
$(n)+(n+1)+(n+2)+(n+3)+(n+4)+(n+5) = 6(n+2)+3$\\
There is always +3 so it will not be divisible by 6 and it is false. It is disproven. \includegraphics[scale=0.70]{billeder/xzibit}\\


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% Problem 6
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\section{Consider the following statements:}
1 = 1,\\
1 + 3 = 4,\\
1 + 3 + 5 = 9,\\
1 + 3 + 5 + 7 = 16,\\
1 + 3 + 5 + 7 + 9 = 25.\\
\begin{enumerate}[(a)]
\item Based on the three statements given above, what is the next statement suggested by these?
1 + 3 + 5 + 7 + 9 + 11 = 36.
\item What conjecture is suggested by these statements?\\
$\forall k \in \mathbb{Z}^{+}$. $1 + 3 + 5 + 7 + 9 + ... + 2k + 1 = (k + 1)^2$. 
\item Verify the conjecture in (b) using induction.
\end{enumerate}
\textbf{Base Case:}\\
Let k be 0: $2*0 + 1 = 1^2 => 1 = 1$ so this holds True.\\
We assume p(k)
$1 + 3 + 5 + 7 + 9 + ... + 2k + 1 = (k + 1)^2$.\\
Then let p(k + 1) : $1+3+5+7+9+...+2k+1 + 2(k + 1) + 1=((k+1)+1)^2$\\
We see that $1 + 3 + 5 + 7 + 9 + ...+2k+1$ can be replaced with $(k+1)^2$\\
Let p(k + 1) : $(k + 1)^2 + 2(k + 1) + 1=((k+1)+1)^2$\\
$((k+1)+1)^2$ = $k^2 + 2k + 1 + 2k + 3$\\
$((k+1)+1)^2$ = $k^2 + 4k + 4$\\
$((k+1)+1)^2$ = $(k + 2)^2$\\
$((k+1)+1)^2$ = $((k + 1) + 1)^2$\\
 So this holds true for our inductions step. Which means that our conjecture is true. \includegraphics[scale=0.70]{billeder/xzibit}


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% Problem 7
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\section{Using induction, prove that}
\begin{enumerate}[(a)]
\item $\forall n \in \mathbb{N}, if n\geq 2, then \ n^3-n$ is always divisable by 3
\item $\forall n \in \mathbb{N},n < 2^n$
\end{enumerate}
\textbf{(a)}\\
Base Case:\\
Let n = 2. $2^3 - 2 = 6$ which is divisable by 3.\\

Assume p(k): $k^3-k = 3m$ where m is an arbitrary integer.\\
Let p(k+1):\\
 = $(k + 1)^3 - (k + 1)$\\
 = $k^3 + 3k^2 + 3k + 1 - k - 1$\\
 = $k^3 - k +3k^2 + 3k$\\
 = $3m +3k^2 + 3k$ because $k^3 - k = 3m$ as established by our p(k)\\
 = $3(m +k^2 + k)$\\
 Which is always divisable by 3. So our induction step holds true. \includegraphics[scale=0.70]{billeder/xzibit}\\
\textbf{(b)}\\
Base Case:\\
Let n = 1. $1<2^1$ which is true.\\

Assume p(k): $k < 2^k$\\
Let p(k+1):\\
 = $(k + 1) < 2^{k + 1}$\\
 = $(k + 1) < 2*2^{k}$\\
 = $(k + 1) < 2^{k} + 2^{k}$\\
 = $1 <  + 2^{k} - k$
 = $1 - 2^{k} < 2^{k} - k$.\\
 = $1 - k < 2^{k} - k$ because $k < 2^k$\\
 = $1 < 2^{k}$
  Which holds true for all natural values of k.\includegraphics[scale=0.70]{billeder/xzibit}

